Question

One particle has a mass of 3.50 x 10-3 kg and a charge of +7.39 μC. A second particle has a mass of 7.65 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.184 m, the speed of the 3.50 x 10-3 kg-particle is 180 m/s. Find the initial separation between the particles.

Answer 1

By Conservation of momentum

$m_{1}V_{1A}+m_{2}V_{2A}=m_{1}V_{1B}+m_{1}V_{2B}$

Here V_1A=V_2A=0

0=(3.5*10^-3)*180 +(7.65*10^-3)*V_2B

V_2B=-82.353 m/s

By Conservation of energy

$\frac{1}{2}m_{1}v_{1A}^{2}+\frac{1}{2}m_{2}v_{2A}^{2}+\frac{Kq_{1}q_{2}}{r_{A}}=\frac{1}{2}m_{1}v_{1B}^{2}+\frac{1}{2}m_{2}v_{2B}^{2}+\frac{Kq_{1}q_{2}}{r_{B}}$

$\frac{Kq_{1}q_{2}}{r_{A}}=\frac{1}{2}m_{1}v_{1B}^{2}+\frac{1}{2}m_{2}v_{2B}^{2}+\frac{Kq_{1}q_{2}}{r_{B}}$

$\frac{(9\times 10^{9})(7.39\times 10^{-6})^{2}}{r_{A}}=$$\frac{(9\times 10^{9})(7.39\times 10^{-6})^{2}}{r_{A}}=\frac{(9\times 10^{9})(7.39\times 10^{-6})^{2}}{0.184}+\frac{1}{2}\times (3.5\times 10^{-3})\times 180^{2}+\frac{1}{2}\times (7.65\times 10^{-3})\times \times (-82.353)^{2}$

$r_{A}=5.76\times 10^{-3}m$